OCR GCE Maths guide - Mathshelper

Wednesday 18 May 2016 – MorningAS GCE MATHEMATICS

4721/01 Core Mathematics 1

QUESTION PAPER

*5974058503*

INSTRUCTIONS TO CANDIDATESThese instructions are the same on the Printed Answer Book and the Question Paper.• The Question Paper will be found inside the Printed Answer Book.• Write your name, centre number and candidate number in the spaces provided on the

Printed Answer Book. Please write clearly and in capital letters.• Write your answer to each question in the space provided in the Printed Answer

Book. If additional space is required, you should use the lined page(s) at the end of the Printed Answer Book. The question number(s) must be clearly shown.

• Use black ink. HB pencil may be used for graphs and diagrams only.• Answer all the questions.• Read each question carefully. Make sure you know what you have to do before starting

your answer.• Do not write in the bar codes.• You are not permitted to use a calculator in this paper.• Give non-exact numerical answers correct to 3 significant figures unless a different

degree of accuracy is specified in the question or is clearly appropriate.

INFORMATION FOR CANDIDATESThis information is the same on the Printed Answer Book and the Question Paper.• The number of marks is given in brackets [ ] at the end of each question or part

question on the Question Paper.• You are reminded of the need for clear presentation in your answers.• The total number of marks for this paper is 72.• The Printed Answer Book consists of 12 pages. The Question Paper consists of 4 pages.

Any blank pages are indicated.

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recycled. Please contact OCR Copyright should you wish to re-use this document.

OCR is an exempt CharityTurn over

© OCR 2016 [Y/102/2693]DC (NH/SW) 123858/3

Candidates answer on the Printed Answer Book.

OCR supplied materials:• Printed Answer Book 4721/01• List of Formulae (MF1)

Other materials required:None

Duration: 1 hour 30 minutes

Oxford Cambridge and RSA

No calculator can be used for this paper

2

4721/01 Jun16© OCR 2016

Answer all the questions.

1 (i) Simplify 2 3 2 3x x2 2- - -^ ^h h . [2]

(ii) Find the coefficient of x3 in the expansion of 3 3 4 5 2x x x x2 3- + - -^ ^h h. [2]

2 Express 3 5

3 20

++ in the form 5a b+ . [4]

3 Solve the simultaneous equations34x y2 2 =+ , 3 4 0x y- =+ . [5]

4 Solve the equation 2 7 3 0y y2

1

4

1

+ =- . [5]

5 Express the following in the form 2p.

(i) 2 25 7 3'^ h [2]

(ii) 5 4 3 163

2

3

1

# #+ [3]

6 (i) Express 4 12 2x x2-+ in the form a x b c2 ++^ h . [4]

(ii) State the coordinates of the maximum point of the curve 4 12 2y x x2= -+ . [2]

7 (i) Sketch the curve 3y x x2= -^ h stating the coordinates of points of intersection with the axes. [3]

(ii) The curve 3y x x2= -^ h is translated by 2 units in the positive direction parallel to the x-axis. State the equation of the curve after it has been translated. [2]

(iii) Describe fully a transformation that transforms the curve 3y x x2= -^ h to 3y x x2

1 2= -^ h. [2]

8 A curve has equation 2y x2= . The points A and B lie on the curve and have x-coordinates 5 and 5 h+ respectively, where h > 0.

(i) Show that the gradient of the line AB is 20 + 2h. [3]

(ii) Explain how the answer to part (i) relates to the gradient of the curve at A. [1]

(iii) The normal to the curve at A meets the y-axis at the point C. Find the y-coordinate of C. [3]

9 Find the set of values of k for which the equation 2 11 2 1x x k x2 = -+ + ^ h has two distinct real roots. [7]

3

4721/01 Jun16© OCR 2016

10y

xAO

The diagram shows the circle with equation 8 6 20 0x y x y2 2+ - - - = .

(i) Find the centre and radius of the circle. [3]

The circle crosses the positive x-axis at the point A.

(ii) Find the equation of the tangent to the circle at A. [6]

(iii) A second tangent to the circle is parallel to the tangent at A. Find the equation of this second tangent. [3]

(iv) Another circle has centre at the origin O and radius r. This circle lies wholly inside the first circle. Find the set of possible values of r. [2]

11 The curve 4 5y x xa2= + + has a stationary point. Find the value of the positive constant a given that the

y-coordinate of the stationary point is 32. [8]

ENDOFQUESTIONPAPER

4721 Mark Scheme June 2016

7

Question Answer Marks Guidance

1 (i) 4x2 – 12x + 9 – 2(9 – 6x + x2) 2x2 – 9

M1 A1 [2]

Square to get at least one 3/4 term quadratic Fully correct www

ISW after correct answer

1 (ii) – 6x3 – 4x3 – 10

B1 B1 [2]

– 6x3 or – 4x3 soi www in these terms Condone – 10x3

Ignore other terms If only embedded in full expansion then award B1B0

2

5353

53203

��

u��

59531

��

M1

B1

A1

Attempt to rationalise the denominator – must attempt to multiply

5220 soi Either numerator or denominator correct and simplified to no more than two terms

Alternative: M1 Correct method to solve simultaneous equations formed from equating expression to ba �5 B1 5220 soi

543

41�� A1

[4] Fully correct and fully simplified. Allow

1 3 54

� � , order reversed etc.

Do not ISW if then multiplied by 4 etc.

A1 Either a or b correct A1 Both correct

3 x2 + (3x + 4)2 = 34 10x2 + 24x – 18 = 0 5x2 + 12x – 9 = 0 (5x – 3)(x + 3) = 0

3,53

� xx

5,529

� yy

M1*

A1

M1dep*

A1

A1 [5]

Substitute for x/y or valid attempt to eliminate one of the variables Correct three term quadratic in solvable form Attempt to solve resulting three term quadratic Correct x values Correct y values

If x eliminated: 10y2 – 8y + 290 = 0 5y2 – 4y + 145 = 0 (5y – 29)(y + 5) = 0 Award A1 A0 for one pair correctly found from correct quadratic Spotted solutions: If M0 DM0

SC B1 529,

53

yx www

SC B1 x = –3, y = –5 www Must show on both line and curve (Can then get 5/5 if both found www and exactly two solutions justified)


8

Question Answer Marks Guidance 4

Let xy 41

2x2 – 7x + 3 = 0 (2x – 1)(x – 3) = 0

21

x , x = 3

4

21¸¹·

¨©§ y , y = 34

161

y , y = 81

Alternative by rearrangement and squaring:

0372 41

21

�� yy , 327 21

41

� yy

912449 21

21

�� yyy , 9437 21

� yy 16y2 – 1297y + 81 = 0 (16 y – 1)( y – 81) = 0

161

y , y = 81

OR methods may be combined:

e.g. after 9437 21

� yy 09374 2

1

�� yy 4x2 – 37x + 9 = 0 (4x – 1)(x – 9) = 0

41

x , x = 9

2

41¸¹·

¨©§ y , y = 92

M1*

M1dep*

A1

M1dep*

A1 [5]

M2*

A1 M1dep*

A1

M1* M1dep*

A1

M1dep* A1 [5]

Use a substitution to obtain a quadratic or

factorise into two brackets each containing 41

y Correct method to solve resulting quadratic Both values correct Attempt to raise to the fourth power Correct final answers Rearrange and square both sides twice Correct quadratic obtained Correct method to solve resulting quadratic Correct final answers Rearrange, square both sides and substitute Correct method to solve resulting quadratic Attempt to square Correct final answers

No marks if whole equation raised to fourth power etc. No marks if straight to formula with no evidence of substitution at start and no raising to fourth power/fourth rooting at end.

No marks if xy 41

and then 2x – 7x2 + 3 = 0. Spotted solutions: If M0 DM0 or M1 DM0 SC B1 y = 81 www SC B1

161

y www

(Can then get 5/5 if both found www and exactly two solutions justified)


9

Question Answer Marks Guidance 5 (i)

(2– 2) 3 or 215 ÷ 221

2– 6

B1

B1 [2]

Valid attempt to simplify Correct answer. Accept p = – 6.

Correct use of either index law

6

21¸¹·

¨©§ oe is B1

5 (ii) � � � �

313

34

31

31

34

34

31

432

2

2

28

262102325

2325

u

u�uu�u

u�u

or

M1

B1

A1 [3]

Attempts to express both terms or a combined term as a power of 2

Correctly obtains 34

2 or 31

2 for either term Correct final answer

e.g. Both 4 = 22 and 16 = 24 soi If M0

SC B1 for 31

168u or 32

48u

6 (i) – 2(x2 – 6x – 2) = – 2[(x – 3) 2 – 2 – 9)] = – 2 (x – 3) 2 + 22

B1 B1 M1 A1 [4]

or a = – 2 b = – 3 4 + 2b2 c = 22 If a, b and c found correctly, then ISW slips in format. If signs of all terms changed at start, can only score SC B1 for fully correct working to obtain 2 (x – 3) 2 – 22 If done correctly and then signs changed at end, do not ISW, award B1B1M1A0

– 2(x – 3)2 – 22 B1 B1 M0 A0 – 2(x – 3) + 22 4/4 (BOD) – 2(x – 3x)2 +22 B1 B0 M1 A0 – 2(x2 – 3)2 + 22 B1 B0 M1 A0 – 2(x + 3)2 + 22 B1 B0 M1 A0 – 2x(x – 3)2 + 22 B0 B1 M1 A0 – 2(x2 – 3) + 22 B1 B0 M1 A0

6 (ii) (3, 22)

B1ft B1ft [2]

Allow follow through “– their b” Allow follow through “their c”

May restart. Follow through marks are for their final answer to (i)


10

Question Answer Marks Guidance 7 (i)

B1

B1

B1

[3]

Negative cubic with a max and a min Cubic that meets y-axis at (0, 0) only Double root at (0,0) and single root at (3, 0) and no other roots

For first mark must clearly be a cubic – must not stop at or before x axis, do not allow straight line sections drawn with a ruler/tending to extra turning points etc. Must not be a finite plot.

7 (ii) � � )5(2 2 xxy �� or � � � �32 223 �� xxy

M1

A1 [2]

Translates curve by +2 or – 2 parallel to the x-axis; must be consistent Fully correct, must have “y =”. ISW expansions

e.g. for M1 � � ))2(3(2 2 �� xx but not � � )23(2 2 �� xx

7 (iii) Stretch Scale factor one-half parallel to the y-axis

B1 B1 [2]

Must use the word “stretch” Must have “factor” or “scale factor”. For “parallel to the y axis” allow “vertically”, “in the y direction”.

Do not accept “in/on/across/up the y axis”. Allow second B1 after “squash” etc. but not after “translate” etc.

8 (i) y1 = 50, y2 = 2(5 + h)2

5)5(50)22050( 2

��

hhh

20 + 2h

B1 M1

A1 [3]

Finds y coordinates at 5 and 5 + h Correct method to find gradient of a line segment; at least 3/4 values correct Fully correct working to give answer AG

Need not be simplified

8 (ii) e.g. “As h tends to zero, the gradient will be 20”

B1 [1]

Indicates understanding of limit See Appendix 2 for examples

e.g. refer to h tending to zero or substitute h = 0 into 20 + 2h to obtain gradient at A

8 (iii) Gradient of normal 201

�

0),5(20150 �� xxy

50¼

B1

M1

A1 [3]

Gradient of line must be numerical negative reciprocal of their gradient at A through their A Correct coordinate in any form e.g.

201005,

4201

Any correct method e.g. labelled diagram.


11

Question Answer Marks Guidance 9 x2 + (2 – 2k) x + 11 + k = 0

(2 – 2k)2 – 4(11 + k) 4k2 – 12k – 40 > 0 k2 – 3k – 10 > 0 (k – 5)(k + 2) k < –2, k > 5

M1* M1dep*

A1

M1dep* A1

M1dep* A1 [7]

Attempt to rearrange to a three-term quadratic Uses b2 – 4ac, involving k and not involving x Correct simplified inequality obtained www Correct method to find roots of 3-term quadratic 5 and – 2 seen as roots b2 – 4ac > 0 and chooses “outside region” Fully correct, strict inequalities.

Each Ms depend on the previous M –2 > k > 5 scores M1A0 Allow “k < –2 or k > 5” for A1 Do not allow “k < –2 and k > 5”

10 (i) Centre of circle (4, 3)

45

450209)3(16)4(

2

22

��

r

ryx

B1 M1

A1 [3]

Correct centre (x ± 4)2 – 42 and (y ± 3)2 – 32 seen (or implied by correct answer)

45 or better www

Or r2 = 42 + 32 + 20 soi ISW after 45

10 (ii) At A, y = 0 so x2 – 8x – 20 = 0 (x – 10)(x + 2) = 0 A = (10, 0)

Gradient of radius = 21

10403

� ��

Gradient of tangent = 2 � �202

1020�

� �xy

xy

M1

A1 M1

B1

M1

A1 [6]

Valid method to find A e.g. put y = 0 and attempt to solve quadratic (allow slips) or Pythagoras’ theorem Correct answer found Attempts to find gradient of radius (3 out of 4 terms correct for their centre, their A) Equation of line through their A, any non-zero gradient Correct answer in any three-term form

Alterative for finding gradient: M1 Attempt at implicit differentiation as evidenced by

dxdyy2 term

A1 06822 ��dxdy

dxdyyx and

substitution of (10, 0) to obtain 2.

10 (iii) A’ = (–2, 6) � �102

226�

� �xy

xy

B1 M1 A1 [3]

Finds the opposite end of the diameter Line through their A’ parallel to their line in (ii) Correct answer in any three-term form

Not through centre of circle

10 (iv) OC = 543 22 � (0 <) r < 45 – 5

M1

A1 [2]

Attempts to find the distance from O to their centre and subtract from their radius Correct inequality, condone ≤

ISW incorrect simplification


12

Question Answer Marks Guidance 11 54 12 �� axxy

28 �� axxdxdy

At stationary point, 08 2 � �axx a = 8x3 oe

When a = 8x3, y = 32 32 = 4x2 + 8x2 + 5

23

x oe

a = 27 OR

54 12 �� axxy 28 �� axx

dxdy

3

12

4275432

xxaaxx

�

��

At stationary point, 08 2 � �axx 0)427(8 23 �� xxxx

23

x oe

a = 27

B1 M1

A1 M1 A1

M1

A1

A1 [8]

B1 M1

A1 M1

A1

M1

A1

A1

ax-1 soi Attempt to differentiate – at least one non-zero term correct Fully correct Sets their derivative to 0 Obtains expression for a in terms of x, or x in terms of a www Substitutes their expression and 32 into equation of the curve to form single variable equation Obtains correct value for x. Allow x =

1227 .

Ignore – 23 given as well.

Obtains correct value for a. Ignore –27 given as well. ax-1 soi Attempt to differentiate – at least one non-zero term correct Fully correct Substitutes 32 into equation of the curve to find expression for a Obtains expression for a in terms of x www Sets derivate to zero and forms single variable equation Obtains correct value for x. Allow x =

1227 .

Ignore – 23 given as well.

Obtains correct value for a. Ignore –27 given as well.

2

3 ax oe, a = 18x oe also fine

or expression for a e.g. 932

a


13

APPENDIX 1

Allocation of method mark for solving a quadratic

e.g. 2x2 – x – 6 = 0 1) If the candidate attempts to solve by factorisation, their attempt when expanded must produce the correct quadratic term and one other correct term (with correct sign): (2x – 3)(x + 2) M1 2x2 and – 6 obtained from expansion (2x – 3)(x + 1) M1 2x2 and – x obtained from expansion (2x + 3)(x + 2) M0 only 2x2 term correct 2) If the candidate attempts to solve by using the formula

a) If the formula is quoted incorrectly then M0.

b) If the formula is quoted correctly then one sign slip is permitted. Substituting the wrong numerical value for a or b or c scores M0

22

624)1(1 2

u�uu��r�

earns M1 (minus sign incorrect at start of formula)

22

624)1(1 2

uuu��r

earns M1 (6 for c instead of −6)

22

624)1(1 2

uuu��r�

M0 (2 sign errors: initial sign and c incorrect)

62

624)1(1 2

�u�uu��r

M0 (2c on the denominator)

Notes – for equations such as 2x2 – x – 6 = 0, then b2 = 12 would be condoned in the discriminant and would not be counted as a sign error. Repeating the sign error for a in both occurrences in the formula would be two sign errors and score M0.

c) If the formula is not quoted at all, substitution must be completely correct to earn the M1


14

3) If the candidate attempts to complete the square, they must get to the “square root stage” involving ±; we are looking for evidence that the candidate knows a quadratic has two solutions!

1649

41

1649

41

06161

412

06212

062

2

2

2

2

r �

¸¹·

¨©§ �

�»»¼

º

««¬

ª�¸

¹·

¨©§ �

�¸¹·

¨©§ �

��

x

x

x

xx

xx

If a candidate makes repeated attempts (e.g. fails to factorise and then tries the formula), mark only what you consider to be their last (complete) attempt.

This is where the M1 is awarded – arithmetical errors may be condoned

provided 41

�x seen or implied


15

APPENDIX 2 – this section contains additional subject specific information Example responses to 8ii h is zero so the gradient is 20 B1 The gradient at A is 20 B0 At A x = 5, h = 0 so gradient equals 20 B1 The gradient at A is 20 so h = 0 B0 As h approaches 0, the gradient of AB approaches 20 which is the gradient of A B1

At A, gradient is 20 so it’s 2h more B0

As h were infinitely small, 20 + 2h is the same as the gradient at A, otherwise it’s greater than the gradient at A B1 20

dxdy

, so it is the gradient of A plus a bit more B0

The smaller h is the closer the gradient of AB is to the gradient of the curve at A B1

2h + 20 = 20 so h = 0 B0

As h tends to zero the gradient gets closer and closer to the actual value B1 They’re getting closer to each other B0 The gradient of AB tends to the gradient of the tangent of the curve as h tends to zero B1

The answer of (i) is converging towards the gradient at A B1

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